Question

Prove that if a>b>0, then (a−b)/a < ln a − ln b < (a−b)/b.

Answer 1

if\[ a>b>0\]then\[\frac{ a-b }{ a }>\ln a -\ln b>\frac{ a-b }{ b }\]

Answer 2

\[a>b>0\implies{a\over b}>1\implies\ln{a\over b}>0\qquad\text{...(1)}\\
a>b>0\implies{b\over a}<1\\
a>b>0\implies a-b>0\implies {a-b\over b}={a\over b}-1>2\qquad\text{...(2)}\\
\rm{\&}\qquad{a-b\over a}=1-{b\over a}<0\qquad\text{...(3)}
\]
from 1, 2, and 3.
QED

Answer 3

corrections with (2) and (3)
it should be the other way...
\[{a-b\over b}<0\\{a-b\over a}>2\]

Answer 4

to be more specific, inequality(1) can be expressed as:
\[{a\over b}>\ln{a\over b}>0\]

Answer 5

but shouldn't\[\frac{ a-b }{ b }=\frac{ a }{ b }-1>0\] if \[\frac{ a }{ b }>1\] and in the same way shouldn't \[\frac{ a-b }{ a }=1-\frac{ b }{ a }>0\] if \[ \frac{ b }{ a }<1\]? @electrokid

Answer 6

@Bomull lol yeah.. But you got the point to preceed.. :)

Answer 7

I'm still stuck with this one, if anyone can give me any pointers. How do I know that\[\frac{ a-b }{ a }>\ln a-\ln b\] and\[\ln a-\ln b>\frac{ a-b }{ b }\] ?

Answer 8

\[\huge alog (\frac{ a }{ b })+b>a\]

Answer 9

\[\huge b*log \frac{ a }{ b }+b<a\]

Answer 10

using the 1st and 2nd equations we cann prove it!!

Answer 11

I don't get it..

Answer 12

oh ok so if \[a*\ln \frac{ a }{ b }<a\] then\[\ln \frac{ a }{ b }+b<\frac{ a }{ a }\]
\[\frac{ a-b }{ a }>\ln \frac{ a }{ b }\]

Answer 13

but where do I pull off \[a*\ln \frac{ a }{ b }+b<a\]...

Answer 14

anyone?

Answer 15

@electrokid @deena123

Answer 16

\[\ln \frac{ a }{ b }+1-\frac{ b }{ a }>0\]
\[\ln \frac{ a }{ b }+\frac{ b }{ a }>-1\]
\[\ln \frac{ a }{ b }>\frac{ b }{ a }-1\]
\[\ln \frac{ a }{ b }>\frac{ b }{ a }-\frac{ a }{ a }\]
\[a*\ln \frac{ a }{ b }>b-a\]
but this sint't going ight

Answer 17

*this isn't going right